Alexăndrel of Moldavia
Prince of Moldavia
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Alexăndrel or Alexandru II (1429 – 25 May 1455), son of Iliaș of Moldavia, was the prince (or voivode) of Moldavia in 1449, from 1452 to 1454, and in 1455.
| Alexăndrel | |
|---|---|
| Prince of Moldavia (1st reign) | |
| Reign | February – 12 October 1449 |
| Predecessor | Petru III |
| Successor | Bogdan II |
| Prince of Moldavia (2nd reign) | |
| Reign | February 1452 – August 1454 |
| Predecessor | Bogdan II |
| Successor | Peter Aaron |
| Prince of Moldavia (3rd reign) | |
| Reign | February – 25 May 1455 |
| Predecessor | Peter Aaron |
| Successor | Peter Aaron |
| Born | 1429 |
| Died | 25 May 1455 (aged 25–26) Bilhorod-Dnistrovskyi |
| Spouse | unmarried |
| Dynasty | Bogdan-Mușat |
| Father | Iliaș of Moldavia |
| Mother | Maria of Halshany |
| Religion | Orthodox |
Life
He preferred the alliance with the Polish–Lithuanian Commonwealth,[1] in contrast with Peter III of Moldavia, who was protégé of John Hunyadi, Governor of Hungary.[2] The influence of Hungary weakened after the Ottomans defeated Hunyadi's army in the second Battle of Kosovo in October 1448.[1] With the support of boyars who preferred an alliance with the Commonwealth, Alexăndrel expelled Peter III from Moldavia and seized the throne[1][2] in February 1449.[3] He confirmed the privileges of the merchants of Brașov.[1] According to the Moldavian-Polish chronicle, Alexăndrel also ceded Chilia (now Kiliya in Ukraine) to Hungary, but two other Moldavian chronicles attribute the same act to his predecessor.[4] In October 1449,[3] Hunyadi's other protégé, Bogdan II broke into Moldavia, forcing Alexăndrel to flee.[5][1]
After Bogdan was murdered, Alexăndrel and Petru Aron divided Moldavia among themselves.[6] Alexăndrel took control of southern Moldavia.[6] He united Moldavia with the support of Hunyadi.[6] He signed a treaty with Hunyadi on 16 February 1453, recognizing him as the protector of Moldavia.[5] Petru Aaron expelled him from Moldavia in March or May 1455.[7]
